Question

Is the vertex of the quadratic function below a maximum or minimum?

Answer 1

A quadratic equation is of the form below:

`ax^2+bx+c=y`

The vertex of the quadratic equation is the value of y where the curve cut the y-axis.

To find the vertex, we would first first find the x-coordinate of the equation using

`x=-(b)/(2a)`

To find the vertex, we would substitute for x in the equation to get y.

Given the quadratic function below

`f(x)=-3x^2-4x+7`

It can be observed that

`a=-3;b=-4,c=7`

The x-coordinate of the vertex is as shown below:

`\begin{gathered} x=-(b)/(2a) \\ x=(--4)/(2(-3)) \\ x=(4)/(-6) \\ x=-(2)/(3) \end{gathered}`

The vertex would be

`\begin{gathered} f(-(2)/(3))=-3(-(2)/(3))^2-4(-(2)/(3))+7 \\ f(-(2)/(3))=-3((4)/(9))+(8)/(3)+7 \\ f(-(2)/(3))=-(4)/(3)+(8)/(3)+7 \\ f(-(2)/(3))=(-4+8)/(3)+7 \\ f(-(2)/(3))=(4)/(3)+7 \\ f(-(2)/(3))=1(1)/(3)+7=8(1)/(3)=(25)/(3) \end{gathered}`

Hence, the vertex of the quadratic function is (-2/3,25/3)

It should be noted that when a is positive, the quadratic graph opens upward and the vertex is minimum but when a is negative, the quadratic curve opens downward, and the vertex would be maximum

The graph of the quafratic function given is shown below

It can be observed that the vertex is maximum